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Given $n$ hermitian (positive-semidefinite) operators $Q_1,\cdots,Q_n$ in finite dimensional Hilbert space (the dimension can be very large), is there a mapping $\phi$ maps $Q_i$ to $P_i$, which preserves inner product, i.e. $\langle P_i, P_j\rangle =\langle Q_i,Q_j\rangle$, and all $P_i$'s are hermitian (positive-semidefinite) operators staying in a smaller dimensional space, say $poly(n)$ ?

Further question, given $n^2$ real numbers $c_{ij}$ $1\leq i,j\leq n$, how to decide if there exist $n$ hermitian (positive-semidefinite) operators $P_1,\cdots,P_n$ satisfying $\langle P_i,P_j\rangle=c_{ij}$? If exists, what is the minimum dimension of operators?

For vectors, the questions above are trivial. I wonder if there are any known results for operators? Thanks.

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Since you are making a distinction between vectors and operators, it is not clear to me which inner product you are considering. –  Martin Argerami Jun 21 '11 at 15:06
Sorry, the inner product for operators are defined to be the trace of the product. –  Penghui Yao Jun 22 '11 at 2:05
it might be more interesting to study the case where the inner products are only approximately preserved.... –  Suvrit Jun 23 '11 at 4:02

2 Answers 2

Treating the linear operators as vectors, wouldn't the Johnson-Lindenstrauss lemma give a low-distortion embedding?

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Let $G$ be the Gram matrix, with $ij$-entry $\langle Q_i,Q_j\rangle$. Then the rank of $G$ is the dimension of the space spanned by the operators $Q_i$. So the answer to your first question is no. (The point is that this all works in any inner product space, changing from "vectors" to "operators" makes no difference.)

For the second question, vectors can be encoded as diagonal operators. So this question would only make sense if you were working with some quite restricted class of operators.

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I just realize the question is trivial for vectors and Hermitian case because $n$ vectors or Hermitian matrices can always be put into $n$-dimensional space preserving inner product. So the more interesting case for me is positive-semidefinite matrices, it is just a convex cone, not vector space. Are the questions non-trivial in this case? –  Penghui Yao Jun 22 '11 at 2:21
This may be far from what you are interested in, but I should mention it: things become non-trivial if you consider operator spaces or operator systems: these are (possibly finite-dimensional) subspaces $W$ of bounded operators on a Hilbert space $B(H)$ together with the norms on all spaces of $n\times n$ matrices over $W$. There are several books on the subject, including books by G. Pisier, E. Effros and J.-J. Ruan as well as V. Paulsen. –  Dima Shlyakhtenko Aug 30 '11 at 17:33

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